"Let $ v_1= \left(\begin{matrix}\frac{-2\sqrt(13)}{13}\\0\\\frac{-3\sqrt(13)}{13}\end{matrix}\right) $ and $ v_2= \left(\begin{matrix}0\\1\\0\end{matrix}\right) $ be two orthonormal vectors.
Following crossproduct is given: $v_1 \times v_3 =v_2$.
Determine $v_3$."
I don't quite get how you are supposed to solve this, and it doesn't make it easier that $v_1$ has such weird coordinates. Can anyone see how you are supposed to do?
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Consider \begin{align*} \vec{v}_1 \times \vec{v}_3&=\vec{v}_2\\ \vec{v}_2 \times (\vec{v}_1 \times \vec{v}_3)&=\vec{v}_2 \times \vec{v}_2\\ (\vec{v}_2 \cdot \vec{v}_3)\vec{v}_1 - (\vec{v}_2 \cdot \vec{v}_1)\vec{v}_3 & =\vec{0}\\ (\vec{v}_2 \cdot \vec{v}_3)\vec{v}_1 - \vec{0} & =\vec{0} && (\because \vec{v}_1 \perp \vec{v}_2) \end{align*} Since $\vec{v}_1 \neq \vec{0}$, so we want $\vec{v}_2 \cdot \vec{v}_3=0$. If $\vec{v}_3=\begin{bmatrix}a\\b\\c\end{bmatrix}$, then we can say $\vec{v}_3=\begin{bmatrix}a\\0\\c\end{bmatrix}$.
If $\vec{v}_3$ was to be orthogonal to $\vec{v}_1$ as well, then you have the extra condition $2a+3c=0$. In which case $\vec{v}_3=a\begin{bmatrix}1\\0\\-\frac{2}{3}\end{bmatrix}$.
First of all, about the weird numbers: $v_1$ has crazy coordinates so that it can be a unit-vector. Note that if we multiply $v_1$ by $\sqrt{13}$, we end up with $(-2,0,-3)$.
Let's start by finding one solution to the equation, and then characterize the rest. In order for the equation $v_1 \times v_3 = v_2$ to make sense, $v_2$ needs to be perpendicular to both $v_1$ and $v_3$. We already know that it's perpendicular to $v_1$, but we need to be careful choosing $v_3$. We also need to make sure that $v_3$ is not parallel to $v_1$.
A nice way to make sure that $v_3$ satisfies all these requirements is to find a $v_3$ that is perpendicular to both $v_1$ and $v_2$, a systematic way to find such a vector is to calculate the cross-product $v_1 \times v_2$. We could also make this computation easier by writing $$ v_1 \times v_2 = [\frac{1}{\sqrt{13}}(-2,0,-3)] \times (0,1,0) = \frac{1}{\sqrt{13}} [(-2,0,-3) \times (0,1,0)]. $$ We could also find a perpendicular vector without the cross product. Note that $v$ will only be perpendicular to $v_2$ if the second coordinate is zero, which means that we're looking for a vector perpendicular to $v_1$ that has a zero in the middle. This turns out to be pretty easy by trial and error.
Let's suppose, however, that we go with the cross-product. We end up with $$ v = \frac{1}{\sqrt{13}}(3,0,-2). $$ Based on our earlier discussion, we can be confident that $v$ has the right direction for our $v_3$, that is, $v_1 \times v$ will definitely be parallel to $v_2$. However, we have to check that things match up exactly: $$ v_1 \times v = [\frac{1}{\sqrt{13}}(-2,0,-3)] \times [\frac{1}{\sqrt{13}}(3,0,-2)] = (0,-1,0). $$ So our $v$ is almost right except that $v_1 \times v$ points in the opposite direction (it does have the right length though!) So, $v_3 = -v$ will solve the equation.
Now what if we want all of the solutions? Let's consider what happens if we plug in $v_3 + w$ for some $w$ instead. We find that $$ v_1 \times (v_3 + w) = v_1 \times v_3 + v_1 \times w = v_2 + v_1 \times w. $$ So, $v_3 + w$ works exactly when $v_1 \times w = 0$. This happens exactly when $w$ is parallel to $v_1$.
We have all of the solutions now: a vector will solve the equation if and only if it has the form $$ v = \frac{1}{\sqrt{13}}(-3,0,2) + t v_1 $$ where $t$ can be any real number.